Question

A pure sample of a compound containing only P and F is analyzed to determine the elemental composition and find the empirical formula. however, this requires two separate analyses. The first converts all the fluorine in the sample to silver fluoride, AgF. A sample weighing 0.3271 g gives 0.6219 g AgF. The second sample, with a mass of 0.3271 g, give 0.8055g Ca3(PO4)2. Report the % composition and the empirical formula of this compound.

Answer 1

1) Percent composition of AgF = (1 * 19) / 126

= 15.0 % F

= (1*107 )/126

= 85.0 % Ag

Empirical formula =

15/ 19 =0.78

85 / 107 =0.79

0.79/ 0.78 = 1

0.78 /0.78 =1

Therefore emprical formula = AgF

2 ) Percent composition of Ca3(PO4)2 =

Ca = ( 3 * 40)/308 = 0.38/100 =38% Ca

P = (2 * 30) /308 = 0.194 =19 .4 % P

O = (8 * 128) / 308 = 0.41 = 41 % O

empirical formula =

38 / 40 (mol wt of Ca) = 0.95

19.4/ 31   (mol wt of P ) = 0.62

41 / 16 (mol wt of O) =2.16

Divide by smallest value

0.95/0.62 = 1.5

0.62 /0.62 = 1

2.16 /0.62 = 3.48 = 3.5

For getting smallest whole number we can multiply by 2

1.5 * 2 = 3

1 *2 = 2

3.5 *2 = 5

Therefore empirical formula = Ca3 P2O5